# Complex Analysis: If f is an odd function, find an expression for f.

Complex Analysis

Suppose that $$f$$ is a holomorphic function in $$\mathbb{C}\setminus\{0\}$$ and satisfies $$\vert f(z) \vert \leq \vert z \vert ^{2} +\frac{1}{\vert z \vert ^{2}}$$

If f is an odd function, find an expression for f.

I thought about calculating the Laurent Serie of f, but I can't calculate Laurent's series explicitly (there is no way to calculate the coefficients), Dai thought about estimating, using the invariance of the circle to know which one cancels out. I cannot develop this question

• Hint: Show that $z^2f(z)$ is a polynomial of degree $\leq 4$ – leoli1 Jan 17 at 14:13
• Which of the terms in the Laurent series must vanish? – Vercassivelaunos Jan 17 at 14:18

The inequality implies that for $$|z|<1$$ $$|z^2f(z)|\le |z|^4+1\le 2\ .$$

It follows from Riemann's theorem that $$z=0$$ is a removable singularity for $$g(z):=z^2f(z)$$. Thus, $$g(z)=\sum_{n=0}^\infty a_nz^n$$ on the punctured complex plane.

The polynomial growth $$|g(z)|\le |z|^4+1$$ implies that $$g$$ is a polynomial (or order at most $$4$$) by Liouville’s theorem

So you can write $$z^2f(z)=a_0+a_1z+a_2z^2+a_3z^3+a_4z^4.$$

The parity of $$f$$ then implies that $$a_0=a_2=a_4=0$$ by the fundamental theorem of algebra: $$g(z)+g(-z)=0$$ implies that $$a_0+a_2z^2+a_4z^4=0$$ for all $$z$$. (But a nonzero polynomial has only finitely many roots.)

• I can't understand the fact that "The parity of f then implies that $a_0 = a_2 = a_4 = 0$ by the fundamental theorem of algebra." Could you explain it better? – User Jan 17 at 17:52
• @User: f is odd implies (by the definition of "odd") that $f(z)+f(-z)=0$, so $g(z)+g(-z)=0$.// Do you know why $a_0+a_2z^2+a_4z^4=0$? – user9464 Jan 17 at 17:56
• yes, because we have to $g(z)+g(-z)=2(a_0+ a_2 z^2 + a_4 z^4 )$. $g(z)+g(-z)=0$, implies $a_0+a_2 z^2+a_4 z^4 =0$. – User Jan 17 at 18:04
• @User: good. Note that this is true for all $z$. Do you know the fundamental theorem of algebra? – user9464 Jan 17 at 18:09
• the theorem will guarantee that $a_0 + a_2 z^2 + a_4 z^4 = 0$ has a root in $\mathbb{C}$ – User Jan 17 at 18:16

One can immediately see that for all $$z\in\Bbb C^\times:=\Bbb C\setminus\{0\},$$ we have $$\left|z^2f(z)\right|=|z|^2|f(z)|\le|z|^4+1.$$

As a result, the Laurent series for $$g(z):=z^2f(z)$$ cannot have any terms of power greater than $$4,$$ for if there were any, then for sufficiently large $$|z|,$$ the above inequality would fail to hold. Moreover, we see that $$g$$ is bounded in the punctured unit disk, so its Laurent series has no terms of negative power, and so is a polynomial of degree no greater than $$4.$$

Since $$f$$ is odd, then $$g(-z)=(-z)^2f(-z)=z^2f(-z)=-z^2f(z)=-g(z),$$ and so $$g$$ is odd. Hence, all terms of the Laurent series for $$g$$ have odd power, and so $$g(z)=az^3+bz$$ for some $$a,b\in\Bbb C,$$ whence $$f(z)=az+\frac{b}{z}.$$

Added: To see why all terms of the Laurent series of $$g$$ must have odd power, suppose $$h$$ is an odd function that is holomorphic in $$\Bbb C^\times.$$ Note that for all $$z\in\Bbb C^\times,$$ we have $$0=h(z)+h(-z).\tag{\heartsuit}$$ Now, write $$h(z)=\sum_{n\in\Bbb Z}b_nz^n,$$ and consider out the expansion for the right-hand side of $$(\heartsuit).$$

• I didn't understand why all terms of the Laurent series for g have odd power – User Jan 17 at 15:29
• I've added some details to my answer to help you get there. – Cameron Buie Jan 17 at 16:04
• What is the need to consider the Laurent Series? Could we consider just one series of power? – User Jan 24 at 1:39
• If we knew that $f$ were defined at $0,$ then we could conclude that such a power series were defined. Equivalently, we could conclude that in the Laurent series $$f(z)=\sum_{n\in\Bbb Z}a_nz^n,$$ we have $a_n=0$ for all $n<0,$ and $z=0$ is a removable singularity of $f.$ – Cameron Buie Jan 24 at 1:50
• So, we use the laurent series because $f$ is not defined in z=0. It's just it ? – User Jan 24 at 1:57